CodeForces 631A-Interview

A. Interview

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Blake is a CEO of a large company called "Blake Technologies". He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.

We define function f(x, l, r) as a bitwise OR of integersx_l, x_l + 1, ..., x_r, wherex_i is thei-th element of the array x. You are given two arrays a andb of length n. You need to determine the maximum value of sumf(a, l, r) + f(b, l, r) among all possible1 ≤ l ≤ r ≤ n.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the length of the arrays.

The second line contains n integers a_i (0 ≤ a_i ≤ 10⁹).

The third line contains n integers b_i (0 ≤ b_i ≤ 10⁹).

Output

Print a single integer — the maximum value of sum f(a, l, r) + f(b, l, r) among all possible1 ≤ l ≤ r ≤ n.

Examples

Input

51 2 4 3 22 3 3 12 1

Output

22

Input

1013 2 7 11 8 4 9 8 5 15 7 18 9 2 3 0 11 8 6

Output

46

Note

Bitwise OR of two non-negative integers a andb is the number c = a OR b, such that each of its digits in binary notation is1 if and only if at least one of a or b have 1 in the corresponding position in binary notation.

In the first sample, one of the optimal answers is l = 2 andr = 4, because f(a, 2, 4) + f(b, 2, 4) = (2OR 4 OR 3) + (3 OR 3 OR 12) = 7 + 15 = 22. Other ways to get maximum value is to choosel = 1 and r = 4,l = 1 and r = 5,l = 2 and r = 4,l = 2 and r = 5,l = 3 and r = 4, orl = 3 and r = 5.

In the second sample, the maximum value is obtained for l = 1 and r = 9.

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int wc[1100],wc1[1100];int xx[1100],xx1[1100];int sum[1111];int main(){int n;while(scanf("%d",&n)!=EOF){int i,j;memset(xx,0,sizeof(xx));memset(xx1,0,sizeof(xx1));for(i=0;i<n;i++){scanf("%d",&wc[i]);if(i==0){xx[i]=wc[0];}if(i>=1){xx[i]=max(wc[i-1],xx[i-1]|wc[i]);//printf("%d\n",xx[i]);}}/*for(i=0;i<n;i++){printf("%d\n",xx[i]);}*/for(i=0;i<n;i++){scanf("%d",&wc1[i]);if(i==0){xx1[i]=wc1[0];}if(i>=1){xx1[i]=max(wc1[i-1],xx1[i-1]|wc1[i]);}}int maxx=-1;for (int i=0;i<n;i++)         {          sum[i]=xx1[i]+xx[i];          if (maxx<sum[i]) maxx=sum[i];      }     printf("%d\n",maxx);}return 0;}